Problem: $g(n) = -4n^{2}-3n+4(f(n))$ $h(t) = -3t+f(t)$ $f(n) = 7n$ $ h(g(1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $g(1)$ . Then we'll know what to plug into the outer function. $g(1) = -4(1^{2})+(-3)(1)+4(f(1))$ To solve for the value of $g$ , we need to solve for the value of $f(1)$ $f(1) = (7)(1)$ $f(1) = 7$ That means $g(1) = -4(1^{2})+(-3)(1)+(4)(7)$ $g(1) = 21$ Now we know that $g(1) = 21$ . Let's solve for $h(g(1))$ , which is $h(21)$ $h(21) = (-3)(21)+f(21)$ To solve for the value of $h$ , we need to solve for the value of $f(21)$ $f(21) = (7)(21)$ $f(21) = 147$ That means $h(21) = (-3)(21)+147$ $h(21) = 84$